Editorial Board - CMS-SMC...Si mest la longueur de la m ediane AD dans un triangle ABC, d emontrer...

Crux Mathematicorum is a problem-solving journal at the secondary and university undergraduate levels,

published online by the Canadian Mathematical Society. Its aim is primarily educational; it is not a research

journal. Online submission:

https://publications.cms.math.ca/cruxbox/

Crux Mathematicorum est une publication de resolution de problemes de niveau secondaire et de premier

cycle universitaire publiee par la Societe mathematique du Canada. Principalement de nature educative,

le Crux n’est pas une revue scientifique. Soumission en ligne:


The Canadian Mathematical Society grants permission to individual readers of this publication to copy articles fortheir own personal use.

c© CANADIAN MATHEMATICAL SOCIETY 2019. ALL RIGHTS RESERVED.

ISSN 1496-4309 (Online)

La Societe mathematique du Canada permet aux lecteurs de reproduire des articles de la presente publication a desfins personnelles uniquement.

c© SOCIETE MATHEMATIQUE DU CANADA 2019 TOUS DROITS RESERVES.

ISSN 1496-4309 (electronique)

Supported by / Soutenu par :

• Intact Financial Corporation

• University of the Fraser Valley

Editorial Board

Editor-in-Chief Kseniya Garaschuk University of the Fraser Valley

MathemAttic Editors John McLoughlin University of New Brunswick

Shawn Godin Cairine Wilson Secondary School

Kelly Paton Quest University Canada

Olympiad Corner Editors Alessandro Ventullo University of Milan

Anamaria Savu University of Alberta

Articles Editor Robert Dawson Saint Mary’s University

Associate Editors Edward Barbeau University of Toronto

Chris Fisher University of Regina

Edward Wang Wilfrid Laurier University

Dennis D. A. Epple Berlin, Germany

Magdalena Georgescu BGU, Be’er Sheva, Israel

Shaun Fallat University of Regina

Chip Curtis Missouri Southern State University

Allen O’Hara University of Western Ontario

Guest Editors Vasile Radu Birchmount Park Collegiate Institute

Aaron Slobodin University of Victoria

Ethan White University of British Columbia

Editor-at-Large Bill Sands University of Calgary

IN THIS ISSUE / DANS CE NUMERO

539 Editorial Kseniya Garaschuk

540 MathemAttic: No. 10

540 Problems: MA46–MA50

542 Solutions: MA21–MA25

546 Problem Solving Vignettes: No. 9 Shawn Godin

550 Teaching Problems: No. 7 John McLoughlin

554 Olympiad Corner: No. 378

554 Problems: OC456–OC460

556 Solutions: OC431–OC435

562 How To Write A Crux Article...Revisited! Robert Dawson

564 Problems: 4491–4500

568 Solutions: 4441–4450

Crux MathematicorumFounding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell

Former Editors / Anciens Redacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer,

Shawn Godin

Crux Mathematicorum

with Mathematical MayhemFormer Editors / Anciens Redacteurs: Bruce L.R. Shawyer, James E. Totten, Vaclav Linek,

Shawn Godin

Editorial /539

EDITORIALThis Volume marks the first year of Crux as an open access publication and weowe a big thank you to Intact Foundation for their support.

Unsurprisingly, as a result of open access, we have been witnessing a continuousincrease in the number of submissions and have been adjusting to the new volumeof materials coming in. With our increased reach, we also decided to bring backmaterials that target wider audiences, which resulted in the creation of a newsection MathemAttic aimed at secondary high school students and teachers. Wewelcome feedback and submissions to the section; you can reach its editors [email protected]. As usual, we are aways looking for article submissionsand I’d like to draw your attention to Robert Dawson’s “How To Write A CruxArticle... Revisited!” in this issue to familiarize yourself with our new expectationsand larger variety of acceptable pieces. We also moved to a shorter problems-to-solutions cycle and, as a result, tightened our submission deadlines (watch out forthose!).

Every year when I look back at the latest completed Volume, I am amazed bythe amount of materials and resources that goes through this publication. I amalso constantly humbled by the expertise and impressed with the flexibility andresponsiveness of my editors. I am immensely grateful for their support. In lightof the increased volume of materials, we will be looking to expand the EditorialBoard, so look out for the CMS call for Crux editors or email me directly [email protected] if you would like to apply to join this outstanding group ofindividuals.

Here is to many more mathematics to be discovered and re-discovered on the pagesof Crux.

Kseniya Garaschuk

Copyright © Canadian Mathematical Society, 2019

[email protected]

[email protected]

540/ MathemAttic

MATHEMATTICNo. 10

The problems featured in this section are intended for students at the secondary schoollevel.

Click here to submit solutions, comments and generalizations to anyproblem in this section.

To facilitate their consideration, solutions should be received by February 15, 2020.

MA46. If both x and y are integers, determine all solutions (x, y) for theequation

(x− 8) · (x− 10) = 2y.

MA47. Let E be any point in rectangle ABCD.

Express x in terms of a, b and d.

MA48. Given any triangle ABC where AD is a median of length m, provethat 4m2 = b2 + c2 + 2bc cosA.

MA49. Given that the perimeters of an equilateral triangle T and a square Sare equal, determine the ratio of the area of the equilateral triangle T to the areaof the square S.

MA50. A family of straight lines is determined by the condition that the sumof the reciprocals of the x and y intercepts is a constant k. Show that all membersof the family are concurrent and find the coordinates of their point of intersection.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Crux Mathematicorum, Vol. 45(10), December 2019


MathemAttic /541

Les problemes proposes dans cette section sont appropries aux etudiants de l’ecole sec-ondaire.

Cliquez ici afin de soumettre vos solutions, commentaires ougeneralisations aux problemes proposes dans cette section.

Pour faciliter l’examen des solutions, nous demandons aux lecteurs de les faire parvenirau plus tard le 15 fevrier 2020.

La redaction souhaite remercier Rolland Gaudet, professeur titulaire a la retraite al’Universite de Saint-Boniface, d’avoir traduit les problemes.

MA46. Determiner toutes les solutions entieres (x, y) a l’equation

(x− 8) · (x− 10) = 2y.

MA47. Soit E un point a l’interieur d’un rectangle ABCD.

Exprimer x en termes de a, b et d.

MA48. Si m est la longueur de la mediane AD dans un triangle ABC,demontrer que 4m2 = b2 + c2 + 2bc cosA.

MA49. Sachant que le perimetre d’un triangle equilateral T et d’un carre Ssont egaux, determiner le ratio de la surface du triangle equilateral T par rapporta la surface du carre S.

MA50. Une famille de lignes droites est definie par la condition que la sommedes reciproques d’interceptions avec l’axe des x et l’axe des y soit une constantek. Demontrer que les lignes de cette famille sont concourantes et trouver lescoordonnees de ce point commun d’intersection.



542/ MathemAttic

MATHEMATTICSOLUTIONS

Statements of the problems in this section originally appear in 2019: 45(5), p. 226–229.

MA21. An equilateral triangle is inscribed into a regular hexagon as shownbelow. If the area of the hexagon is 96, find the area of the inscribed triangle.

Originally Problem 11 of the 2017 Savin contest.

We received 5 solutions. We present the solution by Richard Hess and a graphicalsolution by the editor.

Solution 1, by Richard Hess.

The area of a small equilateral triangle is 1, since there are 96 such triangles in thehexagon. Let the side of each small equilateral triangle be s. Then

√3s2/4 = 1.

Let the coloured equilateral triangle have side length d. From the figure we cansee that there exists a triangle with side lengths d, 5s, and 2s with a 120 angleopposite d. From the cosine law we obtain d2 = (2s)2 + (5s)2 + (2s)(5s) = 39s2.The area of the coloured triangle is then

√3d2/4 = 39

√3s2/4 = 39.

Solution 2, by the editor.

The area of each small triangle is 1. By cutting and rearranging the large triangle(see Figure), we observe that the area of the large triangle is equal to the area of39 small triangles.


MathemAttic /543

MA22. Do there exist three positive integers a, b and c for which both a+b+cand abc are perfect squares? Justify your answer.

Originally Problem 1 of grade 8 of 2018 LXXXI Moscow.

We received 10 solutions. We showcase several of them here.

Among the solutions were the following infinite sequences of numbers (a, b, c):

• (1, 22k+2, 24k+2) with k ∈ N [Sefket Arslanagic];

• ((2mp − 2nq)2, (2mq + 2np)2, (m2 + n2 − p2 − q2)2), with m,n, p, q ∈ N[Lorenzo Benedetti];

• (1, y2, y2), where y satisfies x2 − 2y2 = 1 (Pell’s equation) [Joel Schlosberg];

• (1, 2n2 + 2n, 2n2 + 2n) with n ∈ N [Digby Smith];

• (4x2, 4y2, (x2 + y2 − 1)2), with x, y ∈ N [Konstantin Zelator].

MA23. Integer numbers were placed in squares of a 4 × 4 grid so that thesum in each column and the sum in each row are all equal. Seven of the sixteennumbers are known as shown below, while the rest are hidden.

1 ? ? 2

? 4 5 ?

? 6 7 ?

3 ? ? ?

Show that it is possible to determine at least one of the missing numbers. Is itpossible to determine more than one of the missing numbers?

Originally Problem 2 of Spring Junior A-level of XXXIX Tournament of Towns.

We received 4 solutions, 3 of which were correct. We present the solution by JoelSchlosberg, modified by the editor.

We first show that it is possible to determine at least one of the missing squares.Assign variables to the hidden numbers as follows:

1 a b 2c 4 5 de 6 7 f3 g h i

Then

0 = 1 + a+ b+ 2− [c+ 4 + 5 + d]− [e+ 6 + 7 + f ] + [3 + g + h+ i]

+ [1 + c+ e+ 3]− [a+ 4 + 6 + g]− [b+ 5 + 7 + h] + [2 + d+ f + i]

= 2i− 32,

so the number in the lower-right corner must be 16.


544/ MathemAttic

The following two possibilities show that the values of all of the other hiddensquares can differ while satisfying the given requirements, and so none of themcan be determined without additional information:

1 15 16 214 4 5 1116 6 7 53 9 6 16

1 16 15 216 4 5 914 6 7 73 8 7 16

MA24. You have 5 cards with numbers 3, 4, 5, 6 and 7 written on theirbacks. How many five digit numbers are divisible by 55 with the digits 3, 4, 5, 6,and 7 each appearing once in the number (the card with 6 cannot be rotated tobe used as a 9)?


We received 4 solutions, all of which were correct. We present the joint solu-tion by Sarvin Malekaninejad and Shashwat Mookherjee, completed independently,modified by the editor.

There are 4 numbers which satisfy the conditions. To see this, consider the numbera4a3a2a1a0 which is divisible by 55 with the digits 3, 4, 5, 6, and 7 each appearingonce in the number. By hypothesis, it is clear that a4a3a2a1a0 is divisible by 5and 11. Since a4a3a2a1a0 is divisible by 5, we infer that a0 = 5 (0 not being anavailable digit). On the other hand, since a4a3a2a1a0 is divisible by 11, by thetest of divisibility, we conclude that

(a4 + a2 + 5)− (a3 + a1) ≡ 0 (mod 11).

One can easily check that the numbers 74635, 73645, 64735 and 63745 are all thenumbers that satisfy the conditions.

MA25.

1. Find four consecutive natural numbers such that the first is divisible by 3,the second is divisible by 5, the third is divisible by 7 and the fourth isdivisible by 9.

2. Can you find 100 consecutive natural numbers such that the first is divisibleby 3, the second is divisible by 5, . . . the 100th is divisible by 201?


We received 3 correct and complete submissions. Presented is the one by CorneliuManescu-Avram.

1. We have 159 = 3 · 53, 160 = 5 · 32, 161 = 7 · 23, and 162 = 9 · 18.

2. If

n =lcm(3, 5, 7, . . . , 201) + 3

2,


MathemAttic /545

then n is a positive integer and

n+ k − 1 =lcm(3, 5, 7, . . . , 201) + 2k + 1

2

is divisible by 2k + 1 for 1 ≤ k ≤ 100. Taking an odd multiple of the leastcommon multiple instead we obtain infinitely many n with this property.

Math Quotes

The biologist can push it back to the original protist, and the chemist can push itback to the crystal, but none of them touch the real question of why or how thething began at all. The astronomer goes back untold million of years and endsin gas and emptiness, and then the mathematician sweeps the whole cosmosinto unreality and leaves one with mind as the only thing of which we haveany immediate apprehension. Cogito ergo sum, ergo omnia esse videntur. Allthis bother, and we are no further than Descartes. Have you noticed that theastronomers and mathematicians are much the most cheerful people of the lot?I suppose that perpetually contemplating things on so vast a scale makes themfeel either that it doesn’t matter a hoot anyway, or that anything so large andelaborate must have some sense in it somewhere.

Dorothy L. Sayers with R. Eustace, “The Documents in the Case”, New York:Harper and Row, 1930, p 54.


546/ Problem Solving Vignettes

PROBLEM SOLVINGVIGNETTES

No. 9

Shawn Godin

To Assume or Not to Assume

The problem that follows was the September 2019 problem of the month from theCentre for Education in Mathematics and Computing (CEMC) at the University ofWaterloo. The problem of the month is a new feature for the CEMC and consistsof a problem that is meant to be quite challenging for high school students. Oneproblem appears every month with a hint partway through the month and fullsolution at the end. You can check out the problem of the month and otherCEMC features at https://www.cemc.uwaterloo.ca.

A point (x, y) in the plane is called a lattice point if it has integercoordinates. Points P , Q, and R are distinct lattice points. Prove thatthe measure of ∠PQR cannot be 60.

This problem seems easy enough on the surface but you may be wondering “wheredo I begin?” Sometimes a problem is lacking information and we have to makean assumption in order to solve it. For example, if the problem asked: what isthe next number in the sequence 1, 2, 4, 8, 16, . . . ? In this case, since all we aregiven are the numbers, without context, we can only assume that any pattern wefind continues to hold. Thus, if we see that the numbers given satisfy tn+1 = 2tn,we can assume this continues to hold and the sequence continues 32, 64, 128, . . . .In problems of this sort, we must make an assumption in order to get an answerbecause the next number could be anything. For example, searching The On-Line Encyclopedia of Integer Sequences (https://oeis.org) yields our sequence(A000079: Powers of 2), as well as

• A027423: Number of positive divisors of n! (starting with n = 1): 1, 2, 4,8, 16, 30, . . . .

• A000127: Maximal number of regions obtained by joining n points arounda circle by straight lines. Also number of regions in 4-space formed by n− 1hyperplanes.: 1, 2, 4, 8, 16, 31, . . . .

• A006261: The sum of the first six terms of the n-th row in Pascal’s triangle,i.e.

tn =5∑i=0

Çn

i

å1, 2, 4, 8, 16, 32, 63, . . . .

If you allow the sequence to start 1, 1, 2, . . . there are a few more.


Shawn Godin /547

In other situations people make assumptions that lead to an incorrect solution.For example when asked to draw the smallest number of line segments so that atleast one line segment has passed through each dot in the square array picturedbelow, many people will assume that the line segments have to stay within thesquare defined by the dots and conclude that five segments are needed. If wedidn’t make that assumption, we might have eventually found the four segmentsolution on the right. We can also better the solution to three if “dots” are meantto mean “small circles” and our page is large enough.

So we have seen assumptions that we need to make in order to solve a problem aswell as assumptions that people sometimes make that change the conditions of theoriginal problem and don’t give the desired result. There are other “assumptions”that can be made that can simplify the problem, without changing it. Suppose,for example, we had P (13, 9), Q(9, 7) and R(15, 4) and we wanted to measure theangle ∠PQR. What if we moved each point four units to the left and 2 units downto get P ′(9, 7), Q′(5, 5) and R′(11, 2) – would the angle change?

The angle stays the same because the angle is invariant under rigid transformationslike translations, rotations and reflections. As a result, moving all the points bythe same amount doesn’t change the angle. We can say without loss of generality,assume that Q is at the origin. What this means is that we will assume thatQ is (0, 0). We do this, without loss of generality, because we could have hadQ anywhere, then moved it to the origin to make the numbers easier and theproblem would be unchanged. We can also argue that rotating our coordinatesystem by a multiple of 90 changes the location of the points, but not the size ofthe angle. Finally, notice that if we swap the coordinates of P and R the angle∠PQR remains unchanged. Thus, before we start solving the problem we willmake the following assumption:

Assume, without loss of generality, that Q is at the origin, P (a, b) is in the firstquadrant and R(c, d) is in the first or fourth quadrant such that the slope of QPis greater than the slope of QR.

Since we are only interested in the angle ∠PQR, then if gcd(a, b) = d > 1, thenP ′(ad ,

bd

)is on the segment QP and hence ∠PQR = ∠P ′QR. Similarly with

the coordinates of R. So we can further assume, without loss of generality, thatgcd(a, b) = gcd(c, d) = 1 and we are ready to start our proof.


548/ Problem Solving Vignettes

Solution 1: Construct the triangle PQR and inscribe it in a rectangle whosesides are parallel to the coordinate axes.

P (a, b)

Q(0, 0)

R(c, d)

Thus the side lengths of the rectangle are integer, which means so is its area. Therectangle is partitioned into four triangles. Three of the triangles are right angledwith integer legs, which means the area of these triangles are either an integer orhalf of an integer. This means that twice the area of triangle PQR must be aninteger.

But,

[PQR] =1

2|PQ||QR| sin(∠PQR) =

sin(∠PQR)√a2 + b2)

√c2 + d2

2,

hence, if ∠PQR = 60

2[PQR] =

√3

2

√a2 + b2

√c2 + d2.

Clearly the left side of the equation is an integer, so the right side must be as well.Since we have a factor of

√3 on the right, we need either 3 | (a2+b2) or 3 | (c2+d2)

in order to rationalize the√

3. Since 12 ≡ 22 ≡ 1 (mod 3), the only way to get3 | (a2 + b2) is to have 3 | a and 3 | b, which, since we assumed gcd(a, b) = 1 isimpossible. Similarly, we cannot have 3 | (c2 + d2), so we cannot make the angle60.

Solution 2: As in solution 1, we assume that Q is at the origin, P (a, b) is in thefirst quadrant and the slope of QR is smaller than the slope of PQ. Then let θand φ represent the directed angles between the positive x-axis and QP and QR,respectively. That is, if R is in the first quadrant φ > 0, and if R is in the fourthquadrant φ < 0.

Thus, ∠PQR = θ − φ, but tan θ = ba and tanφ = d

c . Hence

tan(∠PQR) = tan(θ − φ)

=tan(θ)− tan(φ)

1 + tan(θ)(tan(φ)

=ba − d

c

1 + bdac


Shawn Godin /549

which is defined and rational as long as 1 + bdac 6= 0. Given that tan 60 =

√3 is

not rational, and if 1 + bdac = 0, we would get ∠Q = 90, then we conclude that

∠Q 6= 60.

Judicially use assumptions in your proofs where they simplify the process withoutaltering the problem itself. You may want to check out the official solutions to thisproblem which are slight variations on mine (or mine is a variation of theirs). Foryour enjoyment, the current problem of the month, for December, is reproducedbelow. You may want to check the others out.

Let a, b, c, and d be rational numbers and f(x) = ax3 + bx2 + cx+ d.Suppose f(n) is an integer whenever n is an integer and that

1

3n3 − n− 2

3≤ f(n) ≤ 1

3n3 + n2 + 2n+

4

3

for every integer n with the possible exception of n = −2.(a) Show that a = 1

3 .(b) Find f(102019)− f(102019 − 1).


550/ Teaching Problems

TEACHING PROBLEMSNo. 7

John McLoughlin

Magic Triangles as a Starting Point

One of the challenges in teaching mathematics is allowing time for processingwhile ensuring that individuals finishing quickly do not diminish opportunities forsatisfaction and accomplishment of others. For instance, blurting out an answerdiminishes both motivation and satisfaction for others attempting to determine asolution. One avenue to address this concern is to provide problems with multiplesolutions. An elementary example of this is posed here with the challenge being tocreate a magic triangle. Magic triangles have equal sums of numbers along eachof the three sides.

Problem

Using the digits 1, 2, 3, 4, 5, and 6, fill in the circles so that the numbers alongeach side of the triangle add up to the same amount.

The Solving Process

Typically people play with the numbers moving them around until a magic triangleappears, or some frustration perhaps emerges. By that time someone in the classwill have found a magic triangle. The instinct upon finding a solution is to stop,until either another person gets a triangular arrangement with a different magicsum, or a teacher prompts the class suggesting multiple solutions are possible.

Suppose that the first person mentions getting a magic triangle with sums of 11along each side, and then the next person gets a magic sum of 9. As a teacher, myinclination would be to tell all of the students that it is possible to find a magictriangle with a sum of 9 along each side while mentioning that there are moresums that work. Those of you looking for a first solution can work with the totalbeing 9, whereas, others can try to find more solutions.

The pedagogical value of the problem is enhanced as students engage at a levelsuited to their own experience with the problem to that point. Further, the class


John McLoughlin /551

has a common goal of trying to identify the various solutions. We will reach aplace where magic triangles with sums of 9, 10, 11 and 12 have been found. Thisrepresents all of the possible solutions.

The Underlying Mathematics

How do we know that there are no other solutions? Would it be practical for usto find these solutions without being so random in our approach? How might themathematics figure into this problem?

Let us begin to consider the possible sums for the magic triangle using the digitsfrom 1 through 6. First, we note that the sum of these numbers is 21. Secondly,note that each of the numbers will appear in one sum, and those in the verticeswill appear a second time. Hence, the largest possible total of the three sumswould be 21 + (4 + 5 + 6) and likewise, the smallest would be 21 + (1 + 2 + 3).The case with the larger total of 36 corresponds to a magic triangle with a sum of12 along each side, whereas, the smaller total would require sums of 9 along thesides. Hence, the only possible magic sums are 9, 10, 11, and 12.

Consider one of the extreme cases as an example. Suppose we want the magic sumto be 9. Recall that necessitates the placement of 1, 2, and 3 in the vertices. Themiddle numbers are placed accordingly to produce sums of 9, as shown.

1

6 5

2 4 3

Suppose that we had wanted to know whether it would be possible to have a magicsum of 12. (Of course, we now know that it is.) How could we proceed?

Observe that 3× 12 = 36 and 1 + 2 + 3 + 4 + 5 + 6 = 21. Hence, the difference of15 must be the sum of the numbers in the vertices. This can only be if we placethe digits 4, 5, and 6 in the corners. (Note that with a triangle the ordering ofthese does not matter.)

5

1 3

6 2 4

It turns out that the two extreme cases are easier to analyze as they only give usone choice for the set of numbers that must appear in the vertices. Suppose thatwe wanted to consider whether it would be possible to make the magic sum 11.Proceeding as above, we require 3 × 11 − 21 = 12 as the sum of the numbers inthe vertices. This gives three possibilities: (1, 5, 6), (2, 4, 6), and (3, 4, 5). Mixedresults will emerge with efforts to complete the triangles. For example, 5 + 6 = 11


552/ Teaching Problems

already without adding a third number. The (3, 4, 5) arrangement is unworkablesince the side with 3 and 4 will require a second 4 to make 11. Hence, only onemagic triangle results with a sum of 11. It is shown below.

2

3 5

6 1 4

Analysis of the case with the magic sum of 10 is included in the concluding set ofquestions to consider.

Concluding comments

In summary, a problem that is accessible to students in elementary school can serveas a starting point for deeper investigation. The simplicity of the concept allowsfor valuable problem solving qualities such as playfulness, conjecturing and patternseeking at the forefront. The relative absence of any mathematical complicationallows for elaboration and development of the balance between finding a solutionand analyzing a problem. The simplicity of the task brings attention to the processand the mathematics.

For those who want to delve deeper into the idea here, it is worth noting that anydiscussion becomes considerably more advanced by having four numbers alongeach side and using the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9. Suggestions are offeredfor consideration beginning with #3 in the closing set of ideas for consideration.

Ideas to Consider

1. Complete the analysis of the elementary magic triangle problem by consid-ering all possible cases for the vertices in a magic triangle with a sum of 10along each side.

2. The smallest and largest digits from our example are 1 and 6. Their sum is7. Revisit the solutions for each of the magic sums (9, 10, 11, and 12) thatworked. For each solution, create a complementary solution by replacingeach number in the solution with its positive difference from 7. That is, foreach of x = 1, 2, 3, 4, 5, and 6, replace x with (7− x). What do you notice?

3. Extend the ideas from the elementary example discussed to the next levelusing the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. Each magic triangle will have fourdigits on each side in this case. Begin by determining the range of possiblemagic sums. What are the smallest and largest possible magic sums?

4. Select either the largest or smallest possible magic sums from Question 3,and create all possible magic triangles for that sum. Note that simply in-terchanging the middle two numbers (those not in the vertices) along a sidewill not be considered to be creating a new magic triangle.


John McLoughlin /553

5. Choose any other possible magic sum between the extreme values. Completethe analysis for that case and determine how many (if any) magic triangleswith that sum are possible.

6. Observe that the extreme digits of 1 and 9 add to 10. Apply the ideas ofQuestion 2 above to any of the magic triangles you found in questions 4and/or 5. Comment on your findings.

The author welcomes feedback via email [email protected]. Comments from teachersor students would be particularly helpful as we move into the second year of thisfeature, Teaching Problems.


[email protected]

554/ OLYMPIAD CORNER

OLYMPIAD CORNERNo. 378

The problems featured in this section have appeared in a regional or national mathematicalOlympiad.

Click here to submit solutions, comments and generalizations to anyproblem in this section


OC456. Solve the system of equations

(x2 + 1)(x− 1)2 = 2017yz

(y2 + 1)(y − 1)2 = 2017zx

(z2 + 1)(z − 1)2 = 2017xy,

where x ≥ 1, y ≥ 1, z ≥ 1.

OC457. On a blackboard are written the numbers 1!, 2!, 3!, . . . , 2017!. Whatis the smallest among these numbers that should be deleted so that the productof all the remaining numbers is a perfect square?

OC458. Let A be the product of eight consecutive positive integers and let kbe the largest positive integer for which k4 ≤ A. Find the number k knowing thatit is represented in the form 2pm, where p is a prime number and m is a positiveinteger.

OC459. Points P and Q lie respectively on sides AB and AC of a triangleABC such that BP = CQ. Segments BQ and CP intersect at R. The circumcir-cles of triangles BPR and CQR intersect again at point S different from R. Provethat point S lies on the angle bisector ∠BAC.

OC460. Prove that the set of positive integers Z+ can be represented as aunion of five pairwise disjoint subsets with the following property: each 5-tupleof numbers of the form (n, 2n, 3n, 4n, 5n), where n ∈ Z+, contains exactly onenumber from each of these five subsets.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



OLYMPIAD CORNER /555

Les problemes presentes dans cette section ont deja ete presentes dans le cadre d’uneolympiade mathematique regionale ou nationale.

Cliquez ici afin de soumettre vos solutions, commentaires ougeneralisations aux problemes proposes dans cette section.



OC456. Resoudre le systeme d’equations

(x2 + 1)(x− 1)2 = 2017yz

(y2 + 1)(y − 1)2 = 2017zx

(z2 + 1)(z − 1)2 = 2017xy,

ou x ≥ 1, y ≥ 1, z ≥ 1.

OC457. Soient les nombres 1!, 2!, 3!, . . . , 2017!. Lequel parmi ces nombrespourrait etre supprime, de facon a ce que le produit des nombres restants soit uncarre parfait ?

OC458. Soit A le produit de huit entiers positifs consecutifs et soit k le plusgros entier positif tel que k4 ≤ A. Determiner le nombre k, prenant pour acquisqu’il est de la forme 2pm, ou p est un nombre premier et m est un entier positif.

OC459. Les points P et Q se situent, respectivement, sur les cotes ABet AC d’un triangle ABC, de facon a ce que BP = CQ. Les segments BQ etCP intersectent en R. Enfin, les cercles circonscrits des triangles BPR et CQRintersectent de nouveau au point S, distinct de R. Demontrer que le point S setrouve sur la bissectrice de l’angle ∠BAC.

OC460. Demontrer que l’ensemble des entiers positifs Z+ peut etre par-titionne comme reunion de cinq ensembles disjoints avec la propriete suivante :chaque 5-tuple de nombres de la forme (n, 2n, 3n, 4n, 5n), avec n ∈ Z+, contientexactement un nombre de chacun des cinq sous ensembles.




OLYMPIAD CORNERSOLUTIONS


OC431. All natural numbers greater than 1 are coloured with blue or red sothat the sum of every two blue numbers (not necessarily distinct) is blue, and theproduct of every two red ones (not necessarily distinct) is red. It is known thatthe number 1024 is blue. What colour can the number 2017 be?

Originally from Moscow Math Olympiad, Problem 2, Grade 10, Final Round 2017.

We received 7 submissions. We present 2 solutions.

Solution 1, by Kathleen E. Lewis, modified by the editor.

The number 2017 can be red. We know that 1024 is blue, so 2 must be blue,because if 2 were red, every power of 2 would have to be red. Since 2 is blue,and the sum of two blues is blue, then every even number must be blue. But wecannot determine the colours of the odds. Since two colours are used, then oneodd number is red, say a. Assume that a ≥ 5. Since a − (a − 2) = 2, then alsoa−2 must be red and applying the same reasoning to a−2 and so on, we get thatall the odd numbers less than a are red. Now, all the powers an with k = 1, 2, . . .are red and there is a power ak > 2017 which is coloured with red. Using the samereasoning, we conclude that all the odd numbers less than ak are red, therefore2017 is red.

Solution 2, by the Missouri State University Problem Solving Group.

Assuming that both colours must be used (otherwise, every number could becoloured blue), we will show that 2017 must be red.

We will prove, more generally, that (assuming both colours must be used) the onlyvalid colourings are ones in which all the multiples of a given prime number areblue and the rest of the numbers are red. We need a lemma.

Lemma. If gcd(a, b) = 1 and a is blue, then b is red.

Proof. Suppose b were blue. Then every number of the form ka+ `b with integersk, ` ≥ 1 must be blue. It is well known that since gcd(a, b) = 1, every integergreater than or equal to N = (a − 1)(b − 1) is of this form. Therefore everyinteger greater than or equal to N must be blue. Now there must be some numberc that is red (since we are assuming both colours are used) so cm is red for allpositive m. But for m sufficiently large, cm > N and so must be blue. This givesa contradiction, so b must be red.

Suppose the number n is blue. Then at least one of the primes dividing it must beblue (if all of the primes dividing n were red, n would be red). Denote this prime



by p. All the multiples of p must be blue and any non-multiple is relatively primeto p so must be red by the lemma.

In the case at hand, since 1024 is blue, 2 must be blue, so the even numbers areblue and the odd numbers are red hence 2017 is red.

Editor’s Comment. The original problem included the hypothesis that it is knownthat when colouring the numbers both colours were used. Unfortunately, this wasmissed in the translation by the Editor. We apologize for that.

OC432. Find the smallest natural number that is a multiple of 80 such thatyou can rearrange two of its distinct digits and the resulting number will also bea multiple of 80.

Originally from Moscow Math Olympiad, Problem 1, Grade 11, Final Round.

We received 3 submissions, of which 1 was correct and complete. We present thesolution by Oliver Geupel.

A number with the desired property is 1520 because it is a multiple of 80, as wellas the number 5120. We prove that 1520 is the smallest number with the requiredproperty.

The multiples of 80 that are less than 1000 are 80, 160, 240, 320, 400, 480, 560,640, 720, 800, 880, and 960. By inspection, they all do not have the property.Hence, the least number with the desired property is a four digit number 1bc0.

We cannot rearrange the leading 1 with the digit c, because a number that endsin 10 is not divisible by 80. If we can rearrange the leading 1 with the digit b,then the number b1c0 − 1bc0 is a multiple of 80. Hence, 80 | 900(b − 1), that is,4 | b− 1, so that b ≥ 5. By inspection, (b, c) = (5, 0) fails, and (b, c) = (5, 2) yieldsthe solution 1520.

Finally, if we can rearrange the digits b and c, then b and c are even digits suchthat b < c, and the number 1cb0− 1bc0 is a multiple of 80, that is 80 | 90(c− b).Hence, c−b is divisible by 8, which leads to b = 0 and c = 8. But this is impossible,because the number 1080 is not divisible by 80.

The proof is complete.

OC433. Consider an isosceles trapezoid ABCD with bases AD and BC. Acircle ω passing through B and C intersects the side AB and the diagonal BD atpoints X and Y , respectively. The tangent to ω at C intersects the line AD at Z.Prove that the points X, Y , and Z are collinear.

Originally from Moscow Math Olympiad, Problem 2, Grade 9, Final Round 2017.

We received 4 submissions. We present the solution by Oliver Geupel.

The line XY intersects the line AD at a point Z ′ (see figure on the next page). Itis enough to show that Z = Z ′.



We have

∠CY Z ′ = 180 − ∠XY C because X, Y , Z ′ are collinear

= ∠CBX because B, C, Y , X are concyclic

= 180 − ∠ADC because ABCD is an isosceles trapezoid

= ∠CDZ ′ because A, D, Z ′ are collinear.

Hence, the points C, Z ′, D, and Y are concyclic.

The circle ω intersects the side CD at a point X ′. It holds

∠Z ′CD = ∠Z ′Y D because C, Z ′, D, Y are concyclic

= ∠XY B because Y is intersection of BD and XZ ′

= ∠CY X ′ because arcs XB and CX ′ have equal lengths

= ∠ZCD angle between tangent and chord.

Therefore, Z = Z ′.

OC434. The acute isosceles triangle ABC (AB = AC) is inscribed in a circlewith center O. The rays BO and CO intersect the sides AC and AB at the pointsB′ and C ′, respectively. A line l parallel to the line AC passes through point C ′.Prove that the line l is tangent to the circumcircle ω of the triangle B′OC.

Originally from Moscow Math Olympiad, Problem 2, Grade 10, Final Round2017.

We received 4 submissions. We present 2 solutions.

Solution 1, by Oliver Geupel.

Let A′ denote the point of intersection of the lines l and AO. Let ϕ denote thesize of the angle ∠OAC.



Since the triangle AOC is isosceles, we have ∠ACO = ϕ. Also, ∠OA′C ′ =∠A′C ′O = ϕ, because the triangle A′OC ′ is homothetic to the triangle AOC.Moreover, ∠B′A′O = ∠OB′A′ = ϕ, since the triangle A′B′O is the reflection ofthe triangle A′C ′O in the line AO.

Since ∠B′A′O has the same size as the inscribed angle ∠B′CO in ω, it followsthat A′ lies on ω.

The angle between the chord A′O and the line l has the size ϕ, which is identical tothe size of the inscribed angle ∠OB′A′ and, thus, to the size of the angle betweenthe chord A′O and the tangent to ω in A′.

We conclude that l is the tangent to ω in A′.

Solution 2, by Ivko Dimitric.

Let A = ∠BAC. The line q =←→AO is the axis of symmetry of 4ABC. Let D be

the intersection point of l and q. Because of the axial symmetry of the triangleABC about q in which the lines AB,AC and BB′, CC ′ each correspond to theother in the pair, it follows that B′D ‖ AB just as C ′D ‖ AC. That implies

∠B′DO = ∠B′DA = ∠DAB =A

2

and ∠DB′C = ∠BAC = A. Since O is the circumcenter of the triangle ABC then

∠DOC =1

2∠BOC = ∠BAC = A.

Because ∠DB′C = ∠DOC = A, we conclude that the quadrilateral DCB′O iscyclic and its circumcircle is ω, implying

∠B′CO = ∠B′DO = ∠OAB =A

2.



By symmetry, ∠OBC ′ = A/2 and since B′D ‖ AB it follows

∠OCD = ∠OB′D = ∠OBC ′ =A

2.

Thus,

∠B′CD = ∠B′CO + ∠OCD =A

2+A

2= A.

Hence, since ∠B′CD = ∠DB′C, the triangle DCB′ is isosceles with base B′C.The common circumcenter S of4B′OC and4B′DC belongs to the perpendicularbisector s of B′C that is also perpendicular to l ‖ B′C and passes through thevertex D. Since the circumcircle ω of triangle B′OC passes through D and theline l is perpendicular to the radius SD of ω at D, it follows that l is tangent tothe circumcircle of triangle B′OC at point D.

OC435. There are n positive numbers a1, a2, . . . , an written on a blackboard.Under each number ai, Vasya wants to write a number bi ≥ ai so that for everypair of numbers chosen from b1, b2, . . . , bn, the ratio of one of them to the other isan integer. Prove that Vasya can write out the required numbers so that

b1b2 · . . . · bn ≤ 2(n−1)/2a1a2 · . . . · an.

Originally from Moscow Math Olympiad, 4th Problem, Grade 10, Final Round2017.

We received only 1 submission. We present the solution by Oliver Geupel.

There are n positive numbers a1, a2, . . . , an written on a blackboard. Under eachnumber ai, Vasya wants to write a number bi ≥ ai so that for every pair of numberschosen from b1, b2, . . . , bn, the ratio of one of them to the other is an integer. Provethat Vasya can write out the required numbers so that

b1b2 · . . . · bn ≤ 2(n−1)/2a1a2 · . . . · an.

For i, j ∈ 1, 2, . . . , n, let ci,j = dlog2 ai − log2 aje. We prove that one of thefollowing n sequences satisfies the conditions of the problem:

2c1,1 · a1, 2c2,1 · a1, 2c3,1 · a1, . . . , 2cn−1,1 · a1, 2cn,1 · a1,2c1,2 · a2, 2c2,2 · a2, 2c3,2 · a2, . . . , 2cn−1,2 · a2, 2cn,2 · a2,2c1,3 · a3, 2c2,3 · a3, 2c3,3 · a3, . . . , 2cn−1,3 · a3, 2cn,3 · a3,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2c1,n · an, 2c2,n · an, 2c3,n · an, . . . , 2cn−1,n · an, 2cn,n · an.

In the j-th sequence, we have

bi = 2ci,j · aj = 2dlog2 ai−log2 aje · aj ≥ 2log2 ai−log2 aj · aj = ai.



Moreover, the ratio of every two members of the sequence is a power of 2 with aninteger exponent. Hence, the ratio of one of them to the other is an integer.

Note that for real numbers x and y it holds dxe + dye ≤ dx + ye + 1. Thus, fori, j ∈ 1, 2, . . . , n, it holds

ci,j + cj,i = dlog2 ai − log2 aje+ dlog2 aj − log2 aie ≤ d0e+ 1 = 1.

As a consequence, the product of all of the n2 members of the n sequences is

2

(∑n

i,j=1ci,j)· an1an2 · · · ann = 2

(∑1≤i<j≤n

ci,j+cj,i)· an1an2 · · · ann

≤ 2(n−1)n/2 · an1an2 · · · ann.

We conclude that the product of the members of at least one of the n sequencesis not greater than 2(n−1)/2 · a1a2 · · · an. This completes the proof.


562/ How To Write A Crux Article...Revisited!

How To Write A Crux Article...Revisited!Robert Dawson

About four years ago (November 2015, to be precise) I wrote a note for Cruxdetailing what a Crux article should be – and what it should not be. Since then,much has changed. Most noticeably, we’re now an electronic-only journal. Whilethe purple paper problem poser was familiar to many, there are advantages to thenew regime. Not only is the new e-Crux free to all the day that it comes out, butwe have no printing costs. While that doesn’t mean that we can publish everythingpeople send us, it does give us a little more flexibility. Here’s what that means tous – and to you.

My first point in 2015 was that Crux is accessible. “Crux is read by univer-sity professors and graduate students. It’s also read by undergraduates, schoolteachers, school students, and amateurs whose day jobs have nothing to do withmathematics. We ask prospective writers to write for a very clever high schoolstudent. Assume high intelligence but not a lot of specialized knowledge.” Stilltrue! We stop about where the Putnam does – if your article needs much morethan standard sophomore math, it’s not for us.

My second point was that Crux is for problem solvers. “The sort of thing thatmight appear on a regional or national math contest.” That’s still the case, and westill want articles about tricks for problem solving. “Assume that most Crux read-ers know the standard tricks of the trade. Don’t stop and explain mathematicalinduction or double counting unless you’re explaining something new or unusualabout those topics.”

Not everything to do with problem solving makes a good Crux article.

A new problem, with or without solution: send it to Crux as a problem (notan article)

A solution to a Crux problem: submit it to Crux as a solution (not an article)

One problem, many solutions: unlikely to be of much interest.

One technique that works for multiple problems: that’s a Crux article.

I wrote that “Crux is short. Our articles don’t usually run more than five or sixpages, and we’re more likely to run it if it’s three or four. That’s the right lengthfor the sort of thing we publish. We do like well-done illustrations, nice examples,and interesting asides – but please keep it all brief.” This is still true – we have abit more flexibility but our readers’ free time is still the same.

Crux is still not a research journal. If your research paper is high level, sendit to the CMS Bulletin or one of the many other great math journals out there.If you’re an undergraduate student, try the PUMP Journal of Undergraduate Re-search. If it’s elementary, try the College Math Journal, Mathematics Magazine, or


Robert Dawson /563

the American Mathematical Monthly. If it’s accessible and quirky, try the Math-ematical Intelligencer. But don’t send it to us, that’s not what we do. And inparticular, we don’t want “research” that other journals won’t print. “You can-not trisect the angle, square the circle, or duplicate the cube using classical tools.If you understand Wantzel’s classic 1837 proof, you won’t try. If you don’t un-derstand it, you have not done your preparatory work and you have no businesstrying.”

However, with our newfound freedom we have the luxury of running a few general-interest articles. I’m thinking about light and entertaining survey articles, thesort of thing that Martin Gardner used to write. The level will have to be justright – I anticipate rejecting some articles for being too difficult, others for beingtoo elementary, more for being too specialized and still others for being commonknowledge. I suspect that the successful writer will know a lot about the subject,and write much less than they know.

This raises the question of originality. We don’t expect original research – indeed,like Wikipedia, we feel it’s probably not right for us. But we don’t want arti-cles that just rephrase material that’s already available from one source, with orwithout attribution. Your article should combine material from several sources,and say something that none of them says in quite the same way. (Young writ-ers: please run it past a teacher or professor, explaining all your sources, beforesubmitting.)

A note on format. Normally, we’d rather that you use LaTeX. Make sure thatyou know how to use things like theorem environments, \label, \cite, and \ref. Ifyou’re a mathematician you already know LaTeX; if you’re going to be one, you’llhave to learn some day, so might as well do it now. If your article has very fewequations in it, you may submit it as a .doc or .docx file, or even flat text. (PDFis okay but be prepared to submit something editable on request.) Please don’tuse a theorem-proof-lemma form. It’s not our style.

We love diagrams and pictures. Please make them neat; you can create them withMaple, Cinderella, Geogebra, or other software. Now we’re paperless, colour andvideo are both fine, where appropriate. Okay? Now you know what we’re lookingfor. Write it – get somebody to check it over to make sure the math and English(or French) is correct – and send it to us.


564/ Problems

PROBLEMS

Click here to submit problems proposals as well as solutions, commentsand generalizations to any problem in this section.


4491. Proposed by Lorian Saceanu.

Let a, b, c be the side lengths of acute-angled triangle ABC lying opposite of an-gles ∠A,∠B,∠C, respectively. Let r be the inradius of ABC and let R be itscircumradius. Prove that

a∠A+ b∠B + c∠Ca+ b+ c

≤ arccosr

R.

4492. Proposed by George Stoica.

Find the number of classes u in Zn (n ≥ 2) with the property that both u andu− 1 have multiplicative inverses in Zn.

4493. Proposed by Nguyen Viet Hung.

Find all real numbers x, y such thatßx+ 2y + 1

x2 + y2 + 7

™=

1

2

where a denotes the fractional part of a.

4494. Proposed by Michel Bataille.

Let O be the circumcentre of a triangle ABC such that ∠BAC 6= 90 and let γbe the circumcircle of ∆BOC and Ω its centre. If P is a point of the side BC, letQ denote the point of intersection other than O of the line OP and γ. For whichP do the lines OA and ΩQ intersect at M such that MA = MQ?

4495. Proposed by Leonard Giugiuc and Dan Stefan Marinescu.

Prove Mihaileanu’s theorem: Given a point P inside triangle ABC set x = [PBC],y = [PCA], and z = [PAB], where square brackets denote area. If M is a pointon side AB and N is a point on side AC, then line MN contains P if and only if

y · BMMA

+ z · CNNA

= x.

[Ed.: See Solution 2 of 4445.]



Problems /565

4496. Proposed by Leonard Giugiuc.

Let a and b be two fixed numbers such that 0 < a < b. We consider the function

f : [a, b]× [a, b]× [a, b]→ R, f(x, y, z) = (x+ y + z)

Å1

x+

1

y+

1

z

ã.

Find the maximum value of the function.

4497. Proposed by Hoang Le Nhat Tung.

Let a, b, c be positive real numbers. Prove that

b+ c

a+c+ a

b+a+ b

c≥ 4(a2 + b2 + c2)

ab+ bc+ ca+

2(ab+ bc+ ca)

a2 + b2 + c2.

4498. Proposed by Sergey Sadov.

Consider the function f(x) = 1/(x2 + 1) for x > 0. Prove that there exists n suchthat the nth derivative f (n)(x) does not have constant sign for x > 2019.

4499. Proposed by H. A. ShahAli.

Prove that the following system of Diophantine equations has infinitely manyunproportional solutions in positive integers:®

a+ b+ c+ d = e+ f + g,

a2 + b2 + c2 + d2 = e2 + f2 + g2.

4500. Proposed by Chudamani R. Pranesachar.

Let AB be an arc of a circle with radius r and centre O, its angle subtended at thecenter, denoted θ, being less than π. Let M be the mid-point of the shorter arcAB. Points P on radius OA, S on radius OB, Q and R on arc AB are taken suchthat PQRS is a rectangle. Prove that when the area of PQRS is maximum, theline segments OQ, OM , OR divide angle AOB into four equal parts of commonvalue θ

4 ). Determine this maximum area in terms of r and θ.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


566/ Problems

Cliquez ici afin de proposer de nouveaux problemes, de meme que pouroffrir des solutions, commentaires ou generalisations aux problemes

proposs dans cette section.



4491. Proposee par Lorian Saceanu.

Dans un triangle acutangle ABC, soient a, b, c les longueurs ∠A, ∠B, ∠C, re-spectivement. Soient r et R les rayons du cercle inscrit et du cercle circonscrit,repectivement. Demontrer que

a∠A+ b∠B + c∠Ca+ b+ c

≤ arccosr

R.

4492. Proposee par George Stoica.

Determiner le nombre de classes u dans Zn (n ≥ 2) telles que u et u− 1 ont toutesdeux une inverse multiplicative dans Zn.

4493. Proposee par Nguyen Viet Hung.

Determiner tous nombres reels x, y tels queßx+ 2y + 1

x2 + y2 + 7

™=

1

2

ou a denote la partie fractionnelle de a.

4494. Proposee par Michel Bataille.

Soit ABC un triangle tel que ∠BAC 6= 90 et soit O le centre de son cerclecirconscrit ; soit γ le cercle circonscrit de ∆BOC, avec Ω comme centre. Soit Pun point sur le cote BC, la ligne OP intersectant γ en O et Q. Pour quels P leslignes OA et ΩQ intersectent-elles en M de facon a ce que MA = MQ?

4495. Proposee par Leonard Giugiuc et Dan Stefan Marinescu.

Demontrer le theoreme de Mihaileanu : Pour un point P a l’interieur du triangleABC, soient

x = [PBC], y = [PCA], et z = [PAB],



Problems /567

ou les crochets representent les surfaces. Si M est un point sur le cote AB et Nest un point sur le cote AC, alors la ligne MN contient P si et seulement si

y · BMMA

+ z · CNNA

= x.

[Ed.: Voir la solution a 4445.]

4496. Proposee par Leonard Giugiuc.

Soient a et b deux nombres reels tels que 0 < a < b; posons la fonction

f : [a, b]× [a, b]× [a, b]→ R, f(x, y, z) = (x+ y + z)

Å1

x+

1

y+

1

z

ã.

Determiner la valeur maximale de cette fonction.

4497. Proposee par Hoang Le Nhat Tung.

Soient a, b, c des nombres reels positifs. Demontrer que

b+ c

a+c+ a

b+a+ b

c≥ 4(a2 + b2 + c2)

ab+ bc+ ca+

2(ab+ bc+ ca)

a2 + b2 + c2.

4498. Proposee par Sergey Sadov.

Soit la fonction f(x) = 1/(x2 + 1) pour x > 0. Demontrer qu’il existe n tel que lanieme derivee f (n)(x) n’est pas de signe constant pour x > 2019.

4499. Proposee par H. A. ShahAli.

Demontrer qu’il existe infiniment de solutions entieres positives, non proportion-nelles, au systeme Diophantin®

a+ b+ c+ d = e+ f + g,

a2 + b2 + c2 + d2 = e2 + f2 + g2.

4500. Proposee par Chudamani R. Pranesachar.

Soit AB un arc d’un cercle de rayon r et centre O, son angle soutendu au centre,denote θ, etant inferieur a π, epuissoit M le centre de l’arc AB. Les points Psur le rayon OA, Q et R sur l’arc AB et S sur le rayon OB sont choisis pourque PQRS soit un rectangle. Demontrer que lorsque la surface de PQRS estmaximale, les segments OQ, OM et OR divisent ∠AOB en quatre parties egales,de valeur commune θ

4 . Determiner la surface maximale en termes de r et θ.


568/ Solutions

SOLUTIONSNo problem is ever permanently closed. The editor is always pleased to consider forpublication new solutions or new insights on past problems.


4441. Proposed by Mihaela Berindeanu.

Let ABC be an acute triangle, with circumcenter O and orthocenter H. Let A′, B′

and C ′ be the intersection of AH,BH,CH with BC,AC,AB, respectively. LetA1, B1 and C1 be the intersection of AO,BO,CO with BC,AC,AB, respectively.If A′′, B′′ and C ′′ are midpoints of AA1, BB1 and CC1, show that A′A′′, B′B′′

and C ′C ′′ have a common intersection point.

We received 7 submissions, all of which were correct; we feature a solution by IvkoDimitric supplemented by a small, but critical, contribution from the solution byMarie-Nicole Gras.

The requirement that the triangle be acute is unnecessary; we shall see that forany ∆ABC the lines A′A′′, B′B′′, and C ′C ′′all pass through the nine-point center.

When AB = AC, the points A′ and A1 coincide, whence the line A′A′′ is analtitude and passes through the nine-point center.

Let us therefore assume that AB 6= AC, and let M be the midpoint of BC,K the midpoint of HA, and N the intersection point of A′A′′ and KM. SinceAH = 2OM and AH,OM ⊥ BC, we have AK = OM and AK ‖ OM, so thequadrilateral AKMO is a parallelogram and hence KM ‖ AA1. Consequently,triangles NMA′ and A′′A1A

′ are similar as are triangles A′NK and A′A′′A, andwe have

NM

A′′A1=

A′N

A′A′′=NK

A′′A.

Because A′′A1 = A′′A we have NM = NK. Hence, N is the midpoint of thehypotenuse KM of 4A′MK, and therefore the center of its circumcircle.

Since a circle is uniquely determined by any three of its points, and the nine-pointcircle passes through the feet of the altitudes, the midpoints of the sides and themidpoints of the segments from the orthocenter to the vertices, it follows thatthe circumcircle of 4A′MK is the nine-point circle of 4ABC and its center isN = A′A′′ ∩KM. Hence, the line A′A′′ passes through the nine-point center N.

In the same manner, the lines B′B′′ and C ′C ′′ also pass through the nine-pointcenter, so that the three lines in question have the common intersection point N.


Solutions /569

4442. Proposed by Nguyen Viet Hung.

Find the following limit

limn→∞

1√n

Å1√

1 +√

2+

1√3 +√

4+ · · ·+ 1√

2n− 1 +√

2n

ã.

We received 21 submissions, of which all but one were correct. We present severalsolutions.

Solution 1, by Florentin Viescu.

We show that the limit is 1/√

2 as an application of Cesaro-Stolz Lemma. Let

an =1

1 +√

2+

1√3 +√

4+ · · ·+ 1√

2n− 1 +√

2n.

Then

an+1 − an = 1/(√

2n+ 1 +√

2n+ 2).

Let bn =√n. Then bn+1 − bn =

√n+ 1 − √n > 0, for all n ≥ 1. So (bn)n≥1 is

strictly increasing and approaches +∞.

Moreover, the following limit exists

limn→∞

an+1 − anbn+1 − bn

= limn→∞

1√2n+1+

√2n+2√

n+ 1−√n

= limn→∞

√n+ 1 +

√n√

2n+ 1 +√

2n+ 2

= limn→∞

√n

Å»1 + 1

n + 1

ã√n

Å»2 + 1

n +»

2 + 2n

ã=

2

2√

2=

1√2.

Then according to Cesaro-Stolz Lemma limn→∞ an/bn = 1/√

2.

Solution 2, by Nguyen Viet Hung.


2 as an application of the Squeeze Theorem. Wehave clearly

Sn =1√

1 +√

2+

1√3 +√

4+ · · ·+ 1√

2n− 1 +√

2n

<1√

0 +√

1+

1√2 +√

3+ · · ·+ 1√

2n− 2 +√

2n− 1.


570/ Solutions

Therefore,

2Sn < 1 +1√

1 +√

2+

1√2 +√

3+ · · ·+ 1√

2n− 1 +√

2n

= 1 +

√1−√

2

−1+

√2−√

3

−1+ · · ·+

√2n− 1−

√2n

−1

=√

2n. (1)

On the other hand,

Sn =1√

1 +√

2+

1√3 +√

4+ · · ·+ 1√

2n− 1 +√

2n

>1√

2 +√

3+

1√4 +√

5+ · · ·+ 1√

2n+√

2n+ 1.

Therefore,

2Sn >1√

1 +√

2+

1√2 +√

3+ · · ·+ 1√

2n− 1 +√

2n+

1√2n+

√2n+ 1

=

√1−√

2

−1+

√2−√

3

−1+ · · ·+

√2n−

√2n+ 1

−1

=√

2n+ 1− 1. (2)

From (1) and (2), we find that

√2n+ 1− 1

2< Sn <

√2n

2.

We apply Squeeze Theorem to find

limn→∞

Sn√n

=1√2.

Solution 3, by Rob Downes.


2 as an application of the Squeeze Theorem. Let

Sn =1√

1 +√

2+

1√3 +√

4+ · · ·+ 1√

2n− 1 +√

2n

= (√

2− 1) + (√

4−√

3) + · · ·+ (√

2n−√

2n− 1).

Consider f(x) =√

2x−√

2x− 1. Note that f(x) is a strictly decreasing functionfor x ≥ 1. Therefore, as illustrated in the left figure below, we have:

Sn ≥∫ n+1

1

(√

2x−√

2x− 1) dx


Solutions /571

Additionally, as illustrated in the right figure we have:

Sn ≤√

2− 1 +

∫ n

1

(√

2x−√

2x− 1) dx.

Evaluating the integrals, dividing by√n, and simplifying yields the compound

inequality:

(2n+ 2)32 − (2n+ 1)

32 − (2

32 − 1)

3√n

≤ Sn√n

≤ (2n)32 − (2n− 1)

32 − (2

32 − 1) + 3(

√2− 1)

3√n

. (1)

Next, we use the Squeeze Theorem to evaluate the given limit. For the expressionon the left in (1), we have:

limn→∞

(2n+ 2)32 − (2n+ 1)

32 − (2

32 − 1)

3√n

= limn→∞

(2n+ 2)32 − (2n+ 1)

32

3√n

− limn→∞

(232 − 1)

3√n

= limn→∞

12n2 + 18n+ 7

3(√n(2n+ 2)3 +

√n(2n+ 1)3)

=1√2.

Dividing the numerator and denominator by n2 and taking the limit yields theresult 1/

√2. Similarly, for the expression on the right in (1), we have:

limn→∞

(2n)32 − (2n− 1)

32 − (2

32 − 1) + 3(

√2− 1)

3√n

= limn→∞

12n2 − 6n+ 1

3(√

8n4 +√n(2n− 1)3)

=1√2.

Since the limits of the two outer expressions in (1) are equal, we have by theSqueeze Theorem

limn→∞

Sn√n

=1√2.

Editor’s comments. The solutions received used either Cesaro-Stolz Lemma orSqueeze Theorem to calculate the limit. The authors used various inequalities or


572/ Solutions

approximations to be able to apply these theorems. Many solutions were similarand we chose to feature the representative ones.

4443. Proposed by Andrew Wu.

Acute scalene 4ABC has circumcircle Ω and altitudes BE and CF . Point N isthe midpoint of EF and line AN meets Ω again at Z. Let lines ZF and ZE meetΩ again at V and U , respectively, and let lines CV and BU meet at P . Provethat UV and BC meet on the tangent from P to the circumcircle of 4APN .

We received 3 solutions. We present the solution by Andrea Fanchini.

Use barycentric coordinates with reference to 4ABC. We have A(1 : 0 : 0),B(0 : 1 : 0) and C(0 : 0 : 1). Denote the side lengths of the triangle by a, b and c.We use Conway’s notation, in particular S for twice the area of 4ABC, and theshorthand SA, SB and SC , where Sα = S cot(α) for an angle α.

The equation of the circumcircle Ω is a2yz + b2zx + c2xy = 0. As the feet of thealtitudes from B and C respectively, E and F have coordinates E(SC : 0 : SA)and F (SB : SA : 0).

Point N(c2SC + b2SB : b2SA : c2SA) is the midpoint of EF . The line through Aand N has equation c2y − b2z = 0, and meets Ω again at Z(−a2 : 2b2 : 2c2).

The line through Z and F has equation

2c2SAx− 2c2SBy + (a2SA + 2b2SB)z = 0

and meets Ω again at V (2SB(a2SA + 2b2SB) : SA(a2SA + 2b2SB) : −2c2SASB).Similarly, the line through Z and E has equation

2b2SAx+ (a2SA + 2c2SC)y − 2b2SCz = 0


Solutions /573

and meets Ω again at U(2SC(a2SA + 2c2SC) : −2b2SASC : SA(a2SA + 2c2SC)).

The lines CV : SAx − 2SBy = 0 and BU : SAx − 2SCz = 0 meet at the pointP (2SBSC : SASC : SASB).

The circumcircle of 4APN thus has equation

a2yz + b2zx+ c2xy − 4S2 − a2SA4S2 − 2a2SA

(SBy + SCz)(x+ y + z) = 0.

The tangent to this circumcircle at P has equation

t : −2SAS2x+ SB(a2SA + 2c2SC)y + SC(a2SA + 2b2SB)z = 0.

The lines UV : −a2S2Ax + 2SB(a2SA + 2c2SC)y + 2SC(a2SA + 2b2SB)z = 0 and

BC : x = 0 meet at the point X(0 : SC(a2SA + 2b2SB) : −SB(a2SA + 2c2SC)),which we can easily verify is on the tangent t, concluding the proof.

4444. Proposed by Michel Bataille.

Let n be a positive integer. Evaluate in closed form

n−1∑k=0

Åtan2

Å2k + 1

2n+ 1· π

4

ã+ cot2

Å2k + 1

2n+ 1· π

4

ãã.

We received 9 correct solutions. We present the solution by Angel Plaza.

Sincetan2 x+ cot2 x = 4 cot2(2x) + 2 = 4 tan2(π/2− 2x) + 2,

the sum can be written as

4n−1∑k=0

cot2Å

2k + 1

2n+ 1· π

2

ã+ 2n = 4

n∑k=1

tan2

Å2k

2n+ 1· π

2

ã+ 2n.

We will prove that

n∑k=0

tan2(2kπ/2(2n+ 1)) = n(2n+ 1).

Observe that, for 1 ≤ k ≤ n,

1 = (−1)2k =

Åcos

Å2k

2n+ 1· π

2

ã+ i sin

Å2k

2n+ 1· π

2

ãã2n+1

=2n+1∑j=0

Ç2n+ 1

j

åÅcos

Å2k

2n+ 1· π

2

ããj·Åi sin

Å2k

2n+ 1· π

2

ãã2n+1−j.


574/ Solutions

Taking the imaginary parts of both sides and dividing by

cos2n+1

Å2k

2n+ 1· π

2

ã· tan

Å2k

2n+ 1· π

2

ãyields

0 =n∑j=0

Ç2n+ 1

2j

åÅi tan

Å2k

2n+ 1· π

2

ãã2n−2j.

Thus, tan2(2k/(2n+ 1)) with 1 ≤ k ≤ n are the zeros of the polynomial

n∑j=0

Ç2n+ 1

2j

å(−z)n−j .

The sum of these zeros is (2n+1

2

)(2n+1

0

) = n(2n+ 1).

Therefore,n∑k=1

tan2

Å2k

2n+ 1· π

2

ã= n(2n+ 1)

and the proposed sum is

4n(2n+ 1) + 2n = 2n(4n+ 3).

Editor’s comments. Brian Bradie and Oliver Geupel, independently, rendered thesum as

2n∑k=0

tan2

Å2k + 1

2n+ 1· π

4

ã− 1 =

4n+1∑k=1

tan2

Åkπ

8n+ 4

ã−

2n∑k=1

tan2

Åkπ

4n+ 2

ã− 1

=(8n+ 3)(4n+ 1)

3− 2n(4n+ 1)

3− 1

= (2n+ 1)(4n+ 1)− 1 = 8n2 + 6n.

Eight of the solutions relied on the value of a trigonometric sum of the type inthe foregoing solutions. Three of the solvers justified it by a polynomial argumentlike Plaza’s, and one used a partial fractions representation. Two people gave areference to the literature, while the remaining two regarded it as “well-known”and “classical”. The ninth solution appealed to the series development

π2

sin2(πx)=

+∞∑m=−∞

1

(m+ x)2,

and through some delicate bookkeeping got the result from the sum of the oddsquare reciprocals.


Solutions /575

4445. Proposed by Leonard Giugiuc and Dan Stefan Marinescu.

Let ABC be a triangle with AC > BC > AB, incenter I and centroid G .

1. Prove that point A lies in one half-plane of the line GI, while points B andC lie in the other half-plane.

2. The line GI intersects the sides AB and AC at M and N , respectively. Provethat BM = CN if and only if ∠BAC = 60.

We received 5 submissions, all correct. Four of the solutions relied on coordinates;we have chosen one example of such a solution to feature together with the solutionthat avoided coordinates.

Solution 1, by Marie-Nicole Gras.

For triangle ABC we denote by a, b, c the lengths of the sides BC, CA and AB;

by assumption, b > a > c . We put θ =∠BAC

2·

A θ

B

C

I

G

M

N

•

•

The bisectors at vertex A define an orthonormal system with origin A; the cartesiancoordinates of A, B, C are

A(0, 0), B(c cos θ, c sin θ), C(b cos θ,−b sin θ);

using well-known formulas, we get those of G and I :

G

Åb+ c

3cos θ,

c− b3

sin θ

ã, I

Å2bc

a+ b+ ccos θ, 0

ã·

The point M is the intersection of the lines GI and the line AB; the equation ofthe line GI is

c− b3

x sin θ −Åb+ c

3− 2bc

a+ b+ c

ãy cos θ =

2bc

a+ b+ c

c− b3

sin θ cos θ;


576/ Solutions

since the equation of AB is y = x tan θ, we obtainÅc− b

3− b+ c

3+

2bc

a+ b+ c

ãx sin θ =

2bc

a+ b+ c

c− b3

sin θ cos θ;

and we arrive at the coordinates of M :

M

Åc(b− c)a+ b− 2c

cos θ,c(b− c)a+ b− 2c

sin θ

ã·

Exchanging b and c, and the sign of θ, we obtain

N

Åb(c− b)a+ c− 2b

cos θ,b(c− b)a+ c− 2b

sin(−θ)ã·

Since b > a > c we have

c(b− c)a+ b− 2c

=c(b− c)

(a− c) + (b− c) > 0,b(c− b)a+ c− 2b

=b(b− c)

(b− a) + (b− c) > 0,

and then

AM =c(b− c)a+ b− 2c

, AN =b(b− c)

2b− a− c ·

1. Because

MB = AB −AM = c− c(b− c)a+ b− 2c

=c(a− c)a+ b− 2c

> 0,

we deduce that M is between A and B, and

BM =c(a− c)a+ b− 2c

·

Exchanging b and c, we get that N is between A and C, and

CN =b(a− b)a+ c− 2b

=b(b− a)

2b− a− c ·

As points G and I are inside ABC, they belong to the line segment MN , and thenpoint A lies in one half-plane of the line GI, while points B and C lie in the otherhalf-plane. We remark that A is the vertex of ABC such that ∠BAC is between∠ABC and ∠BCA.

2. Finally, we have

BM = CN ⇐⇒ c(a− c)a+ b− 2c

=b(b− a)

2b− a− c⇐⇒ c(a− c)(b− a+ b− c) = b(b− a)(a− c+ b− c)⇐⇒ c(a− c)(b− c)− b(b− a)(b− c) = b(b− a)(a− c)− c(a− c)(b− a)

⇐⇒ (b− c)(ac− c2 − b2 + ab) = (b− c)(ab− bc− a2 + ac)

⇐⇒ (b− c)(b2 + c2 − a2 − bc) = 0.


Solutions /577

By assumption, b 6= c; then BM = CN is equivalent to

a2 = b2 + c2 − bc ⇐⇒ cos(∠BAC) =b2 + c2 − a2

2bc=

1

2⇐⇒ ∠BAC = 60.

Remark. If ∠BAC = 60 and b 6= c, then a2 − c2 = b(b− c) 6= 0 and

CN = BM =c(a− c)(a+ c)

((a− c) + (b− c))(a+ c)=

cb(b− c)b(b− c) + (b− c)(a+ c)

=bc

a+ b+ c·

Let r be the inradius of ABC; then the area of ABC is

a+ b+ c

2r =

1

2bc sin(∠BAC) =

√3

4bc;

we find, finally, that

BM = CN =2r√

3

3=

2

3r tan 60·

Solution 2, by the proposers.

The argument is based on a two-part lemma.

Lemma. Let M be a point on side AB of ∆ABC and N a point on side AC;then

a) the line MN passes through the centroid G if and only if BMMA + CN

NA = 1,and,

b) the line MN passes through the incenter I if and only if b · BMMA + c · CNNA = a.

Editor’s comment: Both parts follow immediately from a theorem of Nicolae Mi-haileanu that deserves to be better known outside Romania. Because its proofwould make a nice challenge for Crux readers, instead of including it here, we haveturned it into one of this month’s problem proposals, problem number 4495.

Back to the problem. We assume that M ,G,I, and N are collinear and set

x =BM

MA, y =

CN

NA. Our lemma tells us that x and y satisfy the simultane-

ous equations, ßx+ y = 1bx+ cy = a

,

which, because b > a > c, gives us

x =a− cb− c > 0 and y =

b− ab− c > 0.

Because there is a unique point, namely M , that divides segment BA in the ratioa− c : b− c, and a unique point, namely N , that divides segment CA in the ratio


578/ Solutions

b−a : b−c, our lemma tells us that with this choice of the points M and N we have

bothBM

MA+CN

NA= 1, so that MN passes through G, and b · BM

MA+ c · CN

NA= a,

so that MN passes through I. We conclude that the line GI separates A fromboth B and C. Furthermore, because

BM

MA=a− cb− c ,

we haveBM

CM +MA=

a− c(a− c) + (b− c) ,

whence

BM =c(a− c)a+ b− 2c

.

similarly,

CN =b(b− a)

2b− a− c .The argument concludes as in part 2 of the first solution.

Editor’s comments. For a compilation of many other properties of triangles withan angle of 60, see Chris Fisher’s article “Recurring Crux Configurations 3: Tri-angles Whose Angles Satisfy 2B = C+A” [37:7 (November 2011) pages 449-453].Problem 4445 is evidently the first Crux problem involving the Nagel line GI ofsuch a triangle.

4446. Proposed by Florin Stanescu.

Let n be a prime number greater than 4 and let A ∈ Mn−1(Q) be such thatAn = In−1. Evaluate det(An−2 + 2An−3 + 3An−4 + · · ·+ (n− 2)A+ (n− 1)In−1)in terms of n.

We received 6 submissions, 5 of which were correct and complete solutions. Wepresent the solution by Brian Bradie, slightly edited.

The condition An = In−1 implies

(A− In−1)(An−1 +An−2 +An−3 + · · ·+A+ In−1) = 0.

Because n is prime, the polynomial

xn−1 + xn−2 + xn−3 + · · ·+ x+ 1

is irreducible over Q. Thus, either A = In−1 or A is similar to the (n−1)× (n−1)matrix

Cn−1 =

0 0 0 · · · 0 −11 0 0 · · · 0 −10 1 0 · · · 0 −10 0 1 · · · 0 −1

. . .

0 0 0 · · · 1 −1

,


Solutions /579

which is the companion matrix associated with the characteristic polynomial

xn−1 + xn−2 + xn−3 + · · ·+ x+ 1.

We now consider two cases.

Case 1 : If A = In−1, then

det(An−2 + 2An−3 + 3An−4 + · · ·+ (n− 2)A+ (n− 1)In−1)

= det

Å(n− 1)n

2In−1

ã=

ï(n− 1)n

2

òn−1.

Case 2 : If A is similar to Cn−1, then

det(An−2 + 2An−3 + 3An−4 + · · ·+ (n− 2)A+ (n− 1)In−1)

= det(Cn−2n−1 + 2Cn−3n−1 + 3Cn−4n−1 + · · ·+ (n− 2)Cn−1 + (n− 1)In−1).

Now note that the companion matrix has the property that

Ckn−1 = (vk+1|vk+2| · · · |vk+n−1),

where vi = vn+i = ei, (i = 1, . . . , n − 1) is the i’th canonical base vector andvn = (−1,−1, . . . ,−1)T .

We obtain

Cn−2n−1 + 2Cn−3n−1 + 3Cn−4n−1 + · · ·+ (n− 2)Cn−1 + (n− 1)In−1

=

n− 1 −1 −1 · · · −1 −1n− 2 n− 2 −2 · · · −2 −2n− 3 n− 3 n− 3 · · · −3 −3

. . .

2 2 2 · · · 2 2− n1 1 1 · · · 1 1

.

Adding j times row n− 1 to row j for j = 1, 2, 3, . . . , n− 2 yields

n 0 0 · · · 0 0n n 0 · · · 0 0n n n · · · 0 0

. . .

n n n · · · n 01 1 1 · · · 1 1

.

Thus,

det(An−2 + 2An−3 + 3An−4 + · · ·+ (n− 2)A+ (n− 1)In−1) = nn−2.


580/ Solutions

4447. Proposed by Lorian Saceanu.

Let ABC be a scalene triangle. Prove that

2 +sinA sinB sinC

sinA+ sinB + sinC≥ sin2A+ sin2B + sin2 C.

We received 19 submissions, all of which are correct. Most of the solutions usedseveral known identities and are very similar to one another. We present the proofby Ioan Viorel Codreanu.

Let r, R, and s denote the inradius, circumradius, and semiperimeter of ∆ABC,respectively. The following identities are all well known:

∏cyc

sinA =sr

2R2,

∑cyc

sinA =s

R,

and ∑cyc

sin2A =s2 − r(4R+ r)

2R2. (1)

Using (1), the proposed inequality is equivalent to

2 +r

2R≥ s2 − r(4R+ r)

2R2, or

s2 ≤ 4R2 + 5Rr + r2, or

s2 ≤ (4R2 + 4Rr + 3r2) + r(R− 2r),

which is true by Gerretsen’s Inequality [Editor’s comment: See for reference, forexample, Item 5.8 on p. 50 of Geometric Inequalities by O. Bottema et al], andR ≥ 2r by Euler’s Inequality.

4448. Proposed by Leonard Giugiuc.

Let a, b, c and d be non-zero complex numbers such that |a| = |b| = |c| = |d| andArg(a) < Arg(b) < Arg(c) < Arg(d). Prove that

|(a− b)(c− d)| = |(a− d)(b− c)| ⇐⇒ (a− b)(c− d) = (a− d)(b− c).

There were 6 correct solutions. Three took the approach of Solution 1 and theother three of Solution 2.

Solution 1, by Oliver Geupel.

The reverse implication is clear. Suppose that |(a− b)(c− d)| = |(a− d)(b− c)|.


Solutions /581

Then for some r, ∣∣∣∣a− bc− b

∣∣∣∣ = r =

∣∣∣∣a− dc− d

∣∣∣∣ ,so that

a− bc− b = reiφ and

a− dc− d = reiψ

for some angles φ and ψ; note that these angles have opposite directions. Since thepoints a, b, c, d in the complex plane form a concyclic quadrilateral φ+ (−ψ) = π.Therefore

a− dc− d = reiψ = rei(φ−π) = −reiφ = −a− b

c− b ,

whence (a− b)(c− d) = (a− d)(b− c).

Solution 2, by Florentin Visescu.

Let a = r(cosm + i sinm), b = r(cosn + i sinn), c = r(cos p + i sin p), d =r(cos q + i sin q). Then, after standard trigonometric manipulations, we find that

(a− b)(c− d) = −4r2 sinn−m

2sin

q − p2

(cos s+ i sin s)

and

(a− d)(b− c) = −4r2 sinq −m

2sin

p− n2

(cos s+ i sin s),

where 2s = m+ n+ p+ q. From this, the desired result follows.

4449. Proposed by Arsalan Wares.

The figure shows two congruent overlapping squares inside a larger square. Thevertices of the overlapping smaller squares divide each of the four sides of thelargest square into three equal parts. If the area of the shaded region is 50, findthe area of the largest square.

We received 18 submissions of which 17 were correct and complete. We presentthe solution by Brian Beasley.


582/ Solutions

(0, 0) (3, 0)

(3, 3)(0, 3)

B

A

(0, 1)

(1, 3)

(3, 2)

(2, 0)

We model the diagram by placing the vertices of the largest square at (0, 0), (0, 3),(3, 3), and (3, 0). Then the shaded region (which we denote by R) is inside thesmaller square with vertices at (0, 1), (1, 3), (3, 2), and (2, 0), yielding an area of5 for the smaller square. Since the smaller square may be partitioned into R andfour remaining congruent triangles, we calculate the area of R by subtracting theareas of the four congruent triangles from 5. By solving for the intersection oflines, it is easy to obtain the coordinates of A and B as (2/3, 2/3) and (1/4, 3/2)respectively. Thus the right-triangle formed by (0, 1), A, and B has area

1

2·√

5

4·√

5

3=

5

24.

Hence in our model, the area of R is 5− 5/6 = 25/6, so in general the ratio of theareas of R and the largest square is (25/6)/9 = 25/54. Thus we conclude that anarea of 50 for R must correspond to an area of 108 for the largest square.

4450. Proposed by Dan Stefan Marinescu, Leonard Giugiuc and Daniel Sitaru.

Let n ≥ 3 be an integer and consider positive real numbers a1, a1, . . . , an such thatan ≥ a1 + a2 + · · ·+ an−1. Prove that

(a1 + a2 + · · ·+ an)

Å1

a1+

1

a2+ · · ·+ 1

an

ã≥ 2((n− 1)2 + 1).

We received 18 solutions, all correct. We present the proof by Angel Plaza, modifiedslightly by the editor.

Let Sn−1 =∑n−1k=1 ak and Tn−1 =

∑n−1k=1

1ak

. Then the proposed inequality readsas

(Sn−1 + an)

ÅTn−1 +

1

an

ã≥ 2((n− 1)2 + 1). (1)


Solutions /583

Now, by the AM-HM inequality, we have

Sn−1Tn−1 =

(n−1∑k=1

ak

)(n−1∑k=1

1

ak

)≥ (n− 1)2,

so

(Sn−1 + an)

ÅTn−1 +

1

an

ã≥ (n− 1)2 + 1 + anTn−1 +

Sn−1an

. (2)

Since an ≥ Sn−1, we have an = Sn−1 + d for some d ≥ 0. Then

anTn−1 +Sn−1an

= (Sn−1 + d)Tn−1 +Sn−1

Sn−1 + d

≥ (n− 1)2 + 1 + dTn−1 +

ÅSn−1

Sn−1 + d− 1

ã= (n− 1)2 + 1 +

dTn−1Sn−1 + d2Tn−1 − dSn−1 + d

≥ (n− 1)2 + 1 +d(n− 1)2 + d2Tn−1 − d

Sn−1 + d≥ (n− 1)2 + 1. (3)

(Therefore d(n− 1)2 − d = d(n2 − 2n) ≥ 0.)

Finally, (1) follows by substituting (3) into (2).

Editor’s comment. Note that equality holds if and only if a1 = a2 = · · · = an−1 = cand an = (n− 1)c for some c > 0.


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